Given a matrix $A$ with a known Jordan decomposition, what is the Jordan decomposition of $A^2+A+I$? So far, I understand that I have to look at each Jordan block. How do I prove that for a Jordan block with a value of x, $J_n(x)$, the Jordan decomposition of $(J_n(x))^2$ is $J_n(x^2)$ ? I have a feeling this will lead me to the solution. Thanks for your time.
 A: Following is a summary of a theorem (Theorem 1.36) from the book Functions of Matrices: Theory and Computation by Nicholas J.Higham.
Let $A$ be a $n \times n$ complex matrix and $f(\lambda)$ be any polynomial.
Let's say $A$ has a Jordan block of size $r$ assoicated with eigenvalue $\lambda$.
We have


*

*If $f'(\lambda) \ne 0$, then $f(A)$ has a corresponding Jordan block of same size $r$.

*If $f'(\lambda) = f''(\lambda) = \cdots f^{(\ell-1)}(\lambda) = 0$ but $f^{(\ell)}(\lambda) \ne 0$ where $\ell \ge 2$, then 


*

*if $\ell \ge r$, the Jordan block of $A$ splits into $r$ copies of $1 \times 1$ Jordan block for $f(A)$ associated with eigenvalue $f(\lambda)$.

*if $\ell \le r-1$, the Jordan block of $A$ splits into following Jordan blocks for $f(A)$:  


*

*$\ell - q$ Jordan blocks of size $p$.  

*$q$ Jordan blocks of size $p+1$.  


where $r = \ell p+ q$ with $0 \le q \le \ell=1, p > 0$.
The book has a proof for this. Since I don't know this stuff, I'm not going to copy it here.
Update
For example, take $f(\lambda) = \lambda^2 + \lambda + 1$. Since $f'(\lambda) = 2\lambda+1$. This means for any Jordan block $J_n(x)$ with $x \ne -\frac12$, $J_n(x)^2 + J_n(x) + I_n$ will be
similar to the Jordan block $J_n( x^2 + x + 1 )$.
A Jordan block $J_n(x)$ (or any matrix similar to it) is characterized by two properties.


*

*$J_n(x) - x I_n$ is nilpotent of order $n$.

*There exists a single vector $v$ such that
$$v,\;J_n(x) v,\;J_n(x)^2 v,\;\ldots,\;J_n(x)^{n-1} v$$
span the whole vector space.


When $f'(x) \ne 0$, one can show following list of vectors from same $v$
$$v,\;f(J_n(x)) v,\;f(J_n(x))^2 v,\;\ldots,\;f(J_n(x))^{n-1} v$$
span the whole vector space. That's why when $f'(x) \ne 0$, $f(J_n(x))$ will be similar to $J_n(f(x))$. I don't know the full details, but Higham's book will definitely cover that.
